EntranceHow to write reaction equations in chemistry 8. Chemical equation
A chemical equation is a recording of a reaction using the symbols of the elements and formulas of the compounds involved in it. The relative amounts of reactants and products, expressed in moles, are indicated by numerical coefficients in the complete (balanced) reaction equation. These coefficients are sometimes called stoichiometric coefficients. Currently, there is an increasing tendency to include indications of the physical states of reactants and products in chemical equations. This is done using the following notations: (gas) or means a gaseous state, (-liquid, ) - a solid, (-aqueous solution.
A chemical equation can be constructed based on experimentally established knowledge of the reactants and products of the reaction being studied, and by measuring the relative amounts of each reactant and product that participate in the reaction.
Writing a Chemical Equation
Writing a complete chemical equation involves the following four steps.
1st stage. Recording the reaction in words. For example,
2nd stage. Replacement of verbal names with formulas of reagents and products.
3rd stage. Balancing the equation (determining its coefficients)
This equation is called balanced or stoichiometric. The need to balance the equation is dictated by the fact that in any reaction the law of conservation of matter must be satisfied. In relation to the reaction we are considering as an example, this means that not a single atom of magnesium, carbon or oxygen can be formed or destroyed in it. In other words, the number of atoms of each element on the left and right sides of a chemical equation must be the same.
4th stage. Indication of the physical condition of each participant in the reaction.
Types of Chemical Equations
Consider the following complete equation:
This equation describes the entire reaction system as a whole. However, the reaction under consideration can also be represented in a simplified form using the ionic equation -.
This equation does not include information about sulfate ions, which are not listed because they do not participate in the reaction under consideration. Such ions are called observer ions.
The reaction between iron and copper(II) is an example of redox reactions (see Chapter 10). It can be divided into two reactions, one of which describes reduction, and the other - oxidation, occurring simultaneously in a general reaction:
These two equations are called half-reaction equations. They are especially often used in electrochemistry to describe processes occurring at electrodes (see Chapter 10).
Interpretation of chemical equations
Consider the following simple stoichiometric equation:
It can be interpreted in two ways. First, according to this equation, one mole of hydrogen molecules reacts with one mole of bromine molecules to form two moles of hydrogen bromide molecules. This interpretation of the chemical equation is sometimes called the molar interpretation.
However, this equation can also be interpreted in such a way that in the resulting reaction (see below) one molecule of hydrogen reacts with one molecule of bromine to form two molecules of hydrogen bromide. This interpretation of a chemical equation is sometimes called its molecular interpretation.
Both molar and molecular interpretations are equally valid. However, it would be completely wrong to conclude, based on the equation of the reaction in question, that one molecule of hydrogen collides with one molecule of bromine to form two molecules of hydrogen bromide. The fact is that this reaction, like most others, is carried out in several successive stages. The set of all these stages is usually called the reaction mechanism (see Chapter 9). In the example we are considering, the reaction includes the following stages:
Thus, the reaction in question is actually a chain reaction involving intermediates called radicals (see Chapter 9). The mechanism of the reaction under consideration also includes other stages and side reactions. Thus, the stoichiometric equation indicates only the resulting reaction. It does not provide information about the reaction mechanism.
Calculation using chemical equations
Chemical equations are the starting point for a wide variety of chemical calculations. Here and later in the book a number of examples of such calculations are given.
Calculation of the mass of reactants and products. We already know that a balanced chemical equation indicates the relative molar amounts of reactants and products involved in a reaction. These quantitative data allow the masses of reactants and products to be calculated.
Let us calculate the mass of silver chloride formed when an excess amount of sodium chloride solution is added to a solution containing 0.1 mol of silver in the form of ions
The first stage of all such calculations is to write the equation of the reaction in question: I
Since the reaction uses an excess amount of chloride ions, it can be assumed that all ions present in the solution are converted into The reaction equation shows that one mole of ions is obtained from one mole. This allows us to calculate the mass of the product as follows:
Hence,
Since g/mol, then
Determination of the concentration of solutions. Calculations based on stoichiometric equations underlie quantitative chemical analysis. As an example, consider determining the concentration of a solution based on the known mass of the product formed in the reaction. This type of quantitative chemical analysis is called gravimetric analysis.
A quantity of potassium iodide solution was added to the nitrate solution, which is sufficient to precipitate all the lead in the form of iodide. The mass of the iodide formed was 2.305 g. The volume of the initial nitrate solution was equal to. It is required to determine the concentration of the initial nitrate solution
We have already encountered the equation for the reaction in question:
This equation shows that one mole of lead(II) nitrate is required to produce one mole of iodide. Let us determine the molar amount of lead (II) iodide formed in the reaction. Because the
Class: 8
Presentation for the lesson
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The purpose of the lesson: help students develop knowledge of a chemical equation as a conditional recording of a chemical reaction using chemical formulas.
Tasks:
Educational:
- systematize previously studied material;
- teach the ability to write equations chemical reactions.
Educational:
- develop communication skills (work in pairs, ability to listen and hear).
Educational:
- develop educational and organizational skills aimed at accomplishing the task;
- develop analytical thinking skills.
Lesson type: combined.
Equipment: computer, multimedia projector, screen, assessment sheets, reflection card, “set of chemical symbols”, notebook with printed base, reagents: sodium hydroxide, iron(III) chloride, alcohol lamp, holder, matches, Whatman paper, multi-colored chemical symbols.
Lesson presentation (Appendix 3)
Lesson structure.
I. Organizing time.
II. Updating knowledge and skills.
III. Motivation and goal setting.
IV. Learning new material:
4.1 combustion reaction of aluminum in oxygen;
4.2 decomposition reaction of iron (III) hydroxide;
4.3 algorithm for arranging coefficients;
4.4 minutes of relaxation;
4.5 set the coefficients;
V. Consolidation of acquired knowledge.
VI. Summing up the lesson and grading.
VII. Homework.
VIII. Final words from the teacher.
During the classes
Chemical nature of a complex particle
determined by the nature of elementary
components,
their number and
chemical structure.
D.I.Mendeleev
Teacher. Hello guys. Sit down.
Please note: you have a printed notebook on your desk. (Appendix 2), in which you will work today, and a score sheet in which you will record your achievements, sign it.
Updating knowledge and skills.
Teacher. We got acquainted with physical and chemical phenomena, chemical reactions and signs of their occurrence. Studied the law conservation of mass substances.
Let's test your knowledge. I suggest you open your printed notebooks and complete task 1. You are given 5 minutes to complete the task.
Test on the topic “Physical and chemical phenomena. Law of conservation of mass of substances.”
1. How do chemical reactions differ from physical phenomena?
- Change in shape and state of aggregation of a substance.
- Formation of new substances.
- Change of location.
2. What are the signs of a chemical reaction?
- Precipitate formation, color change, gas evolution.
3. In accordance with what law are equations of chemical reactions drawn up?
- The law of constancy of the composition of matter.
- Law of conservation of mass of matter.
- Periodic law.
- Law of dynamics.
- The law of universal gravitation.
4. The law of conservation of mass of matter discovered:
- DI. Mendeleev.
- C. Darwin.
- M.V. Lomonosov.
- I. Newton.
- A.I. Butlerov.
5. A chemical equation is called:
- Conventional notation of a chemical reaction.
Teacher. You've done the job. I suggest you check it out. Exchange notebooks and check each other. Attention to the screen. For each correct answer - 1 point. Enter the total number of points on the evaluation sheets.
Motivation and goal setting.
Teacher. Using this knowledge, today we will draw up equations of chemical reactions, revealing the problem “Is the law of conservation of mass of substances the basis for drawing up equations of chemical reactions”
Learning new material.
Teacher. We are accustomed to thinking that an equation is a mathematical example where there is an unknown, and this unknown needs to be calculated. But in chemical equations there is usually nothing unknown: everything is simply written down in them using formulas: which substances react and which are obtained during this reaction. Let's see the experience.
(Reaction of sulfur and iron compound.) Appendix 3
Teacher. From the point of view of the mass of substances, the reaction equation for the compound of iron and sulfur is understood as follows
Iron + sulfur → iron (II) sulfide (task 2 tpo)
But in chemistry, words are reflected by chemical signs. Write this equation using chemical symbols.
Fe + S → FeS
(One student writes on the board, the rest in TVET.)
Teacher. Now read it.
Students. An iron molecule interacts with a sulfur molecule to produce one molecule of iron (II) sulfide.
Teacher. In this reaction, we see that the amount of starting substances is equal to the amount of substances in the reaction product.
We must always remember that when composing reaction equations, not a single atom should be lost or unexpectedly appear. Therefore, sometimes, having written all the formulas in the reaction equation, you have to equalize the number of atoms in each part of the equation - set the coefficients. Let's see another experiment
(Combustion of aluminum in oxygen.) Appendix 4
Teacher. Let's write the equation of a chemical reaction (task 3 in TPO)
Al + O 2 → Al +3 O -2
To write the oxide formula correctly, remember that
Students. Oxygen in oxides has an oxidation state of -2, aluminum is a chemical element with a constant oxidation state of +3. LCM = 6
Al + O 2 → Al 2 O 3
Teacher. We see that 1 aluminum atom enters into the reaction, two aluminum atoms are formed. Two oxygen atoms enter, three oxygen atoms are formed.
Simple and beautiful, but disrespectful to the law of conservation of mass of substances - it is different before and after the reaction.
Therefore, we need to arrange the coefficients in this chemical reaction equation. To do this, let's find the LCM for oxygen.
Students. LCM = 6
Teacher. We put coefficients in front of the formulas for oxygen and aluminum oxide so that the number of oxygen atoms on the left and right is equal to 6.
Al + 3 O 2 → 2 Al 2 O 3
Teacher. Now we find that as a result of the reaction, four aluminum atoms are formed. Therefore, in front of the aluminum atom on the left side we put a coefficient of 4
Al + 3O 2 → 2Al 2 O 3Let us once again count all the atoms before and after the reaction. We bet equal.
4Al + 3O 2 _ = 2 Al 2 O 3
Teacher. Let's look at another example
(The teacher demonstrates an experiment on the decomposition of iron (III) hydroxide.)
Fe(OH) 3 → Fe 2 O 3 + H 2 O
Teacher. Let's arrange the coefficients. One iron atom reacts and two iron atoms are formed. Therefore, before the formula of iron hydroxide (3) we put a coefficient of 2.
Fe(OH) 3 → Fe 2 O 3 + H 2 OTeacher. We find that 6 hydrogen atoms enter into the reaction (2x3), 2 hydrogen atoms are formed.
Students. NOC =6. 6/2 = 3. Therefore, we set the coefficient of 3 for the water formula
2Fe(OH) 3 → Fe 2 O 3 + 3 H 2 O
Teacher. We count oxygen.
Students. Left – 2x3 =6; on the right – 3+3 = 6
Students. The number of oxygen atoms that entered into the reaction is equal to the number of oxygen atoms formed during the reaction. You can bet equally.
2Fe(OH) 3 = Fe 2 O 3 +3 H 2 O
Teacher. Now let's summarize everything that was said earlier and get acquainted with the algorithm for arranging coefficients in the equations of chemical reactions.
- Count the number of atoms of each element on the right and left sides of the chemical reaction equation.
- Determine which element has a changing number of atoms and find the LCM.
- Divide the NOC into indices to obtain coefficients. Place them before the formulas.
- Recalculate the number of atoms and repeat the action if necessary.
- The last thing to check is the number of oxygen atoms.
Teacher. You've worked hard and you're probably tired. I suggest you relax, close your eyes and remember some pleasant moments in life. They are different for each of you. Now open your eyes and make circular movements with them, first clockwise, then counterclockwise. Now move your eyes intensively horizontally: right - left, and vertically: up - down.
Now let’s activate our mental activity and massage our earlobes.
Teacher. We continue to work.
In printed notebooks we will complete task 5. You will work in pairs. You need to place the coefficients in the equations of chemical reactions. You are given 10 minutes to complete the task.
- P + Cl 2 →PCl 5
- Na + S → Na 2 S
- HCl + Mg →MgCl 2 + H 2
- N 2 + H 2 →NH 3
- H 2 O → H 2 + O 2
Teacher. Let's check the completion of the task ( the teacher questions and displays the correct answers on the slide). For each correctly set coefficient - 1 point.
You completed the task. Well done!
Teacher. Now let's get back to our problem.
Guys, what do you think, is the law of conservation of mass of substances the basis for drawing up equations of chemical reactions?
Students. Yes, during the lesson we proved that the law of conservation of mass of substances is the basis for drawing up equations of chemical reactions.
Consolidation of knowledge.
Teacher. We have studied all the main issues. Now let's do a short test that will allow you to see how you have mastered the topic. You should only answer “yes” or “no”. You have 3 minutes to work.
Statements.
- In the reaction Ca + Cl 2 → CaCl 2, coefficients are not needed.(Yes)
- In the reaction Zn + HCl → ZnCl 2 + H 2, the coefficient for zinc is 2. (No)
- In the reaction Ca + O 2 → CaO, the coefficient for calcium oxide is 2.(Yes)
- In the reaction CH 4 → C + H 2 no coefficients are needed.(No)
- In the reaction CuO + H 2 → Cu + H 2 O, the coefficient for copper is 2. (No)
- In the reaction C + O 2 → CO, a coefficient of 2 must be assigned to both carbon monoxide (II) and carbon. (Yes)
- In the reaction CuCl 2 + Fe → Cu + FeCl 2 no coefficients are needed.(Yes)
Teacher. Let's check the progress of the work. For each correct answer - 1 point.
Lesson summary.
Teacher. You did a good job. Now calculate the total number of points scored for the lesson and give yourself a grade according to the rating that you see on the screen. Give me your evaluation sheets so that you can enter your grade into the journal.
Homework.
Teacher. Our lesson came to an end, during which we were able to prove that the law of conservation of mass of substances is the basis for composing reaction equations, and we learned how to compose chemical reaction equations. And as a final point, write down homework
§ 27, ex. 1 – for those who received a rating of “3”
ex. 2 – for those who received a rating of “4”
ex. 3 – for those who received a rating“5”
Final words from the teacher.
Teacher. I thank you for the lesson. But before you leave the office, pay attention to the table (the teacher points to a piece of Whatman paper with an image of a table and multi-colored chemical symbols). You see chemical signs of different colors. Each color symbolizes your mood.. I suggest you make your own table chemical elements(it will differ from D.I. Mendeleev’s PSHE) - a table of the mood of the lesson. To do this, you must go to the sheet of music, take one chemical element, according to the characteristic that you see on the screen, and attach it to a table cell. I will do this first by showing you how comfortable I am working with you.
F I felt comfortable in the lesson, I received answers to all my questions.
F I was bored in class, I didn’t learn anything new.
9.1. What are the chemical reactions?
Let us remember that we call any chemical phenomena in nature chemical reactions. During a chemical reaction, some break down and others form. chemical bonds. As a result of the reaction, other substances are obtained from some chemical substances (see Chapter 1).
While doing your homework for § 2.5, you became acquainted with the traditional selection of four main types of reactions from the entire set of chemical transformations, and then you also proposed their names: reactions of combination, decomposition, substitution and exchange.
Examples of compound reactions:
C + O 2 = CO 2; (1)
Na 2 O + CO 2 = Na 2 CO 3; (2)
NH 3 + CO 2 + H 2 O = NH 4 HCO 3. (3)
Examples of decomposition reactions:
2Ag 2 O 4Ag + O 2; (4)
CaCO 3 CaO + CO 2; (5)
(NH 4) 2 Cr 2 O 7 N 2 + Cr 2 O 3 + 4H 2 O. (6)
Examples of substitution reactions:
CuSO 4 + Fe = FeSO 4 + Cu; (7)
2NaI + Cl 2 = 2NaCl + I 2; (8)
CaCO 3 + SiO 2 = CaSiO 3 + CO 2. (9)
| Exchange reactions- chemical reactions in which the starting substances seem to exchange their components. |
Examples of exchange reactions:
Ba(OH) 2 + H 2 SO 4 = BaSO 4 + 2H 2 O; (10)
HCl + KNO 2 = KCl + HNO 2; (eleven)
AgNO 3 + NaCl = AgCl + NaNO 3. (12)
The traditional classification of chemical reactions does not cover all their diversity - in addition to the four main types of reactions, there are also many more complex reactions.
The identification of two other types of chemical reactions is based on the participation in them of two important non-chemical particles: electron and proton.
During some reactions, complete or partial transfer of electrons from one atom to another occurs. In this case, the oxidation states of the atoms of the elements that make up the starting substances change; of the examples given, these are reactions 1, 4, 6, 7 and 8. These reactions are called redox.
In another group of reactions, a hydrogen ion (H +), that is, a proton, passes from one reacting particle to another. Such reactions are called acid-base reactions or proton transfer reactions.
Among the examples given, such reactions are reactions 3, 10 and 11. By analogy with these reactions, redox reactions are sometimes called electron transfer reactions. You will become acquainted with OVR in § 2, and with KOR in the following chapters.
COMPOUNDING REACTIONS, DECOMPOSITION REACTIONS, SUBSTITUTION REACTIONS, EXCHANGE REACTIONS, REDOX REACTIONS, ACID-BASE REACTIONS.
Write down reaction equations corresponding to the following schemes:
a) HgO Hg + O 2 ( t); b) Li 2 O + SO 2 Li 2 SO 3; c) Cu(OH) 2 CuO + H 2 O ( t);
d) Al + I 2 AlI 3; e) CuCl 2 + Fe FeCl 2 + Cu; e) Mg + H 3 PO 4 Mg 3 (PO 4) 2 + H 2 ;
g) Al + O 2 Al 2 O 3 ( t); i) KClO 3 + P P 2 O 5 + KCl ( t); j) CuSO 4 + Al Al 2 (SO 4) 3 + Cu;
l) Fe + Cl 2 FeCl 3 ( t); m) NH 3 + O 2 N 2 + H 2 O ( t); m) H 2 SO 4 + CuO CuSO 4 + H 2 O.
Indicate the traditional type of reaction. Label redox and acid-base reactions. In redox reactions, indicate which atoms of elements change their oxidation states.
9.2. Redox reactions
Let's consider the redox reaction that occurs in blast furnaces during the industrial production of iron (more precisely, cast iron) from iron ore:
Fe 2 O 3 + 3CO = 2Fe + 3CO 2.
Let us determine the oxidation states of the atoms that make up both the starting substances and the reaction products
| Fe2O3 | + | = | 2Fe | + |
As you can see, the oxidation state of carbon atoms increased as a result of the reaction, the oxidation state of iron atoms decreased, and the oxidation state of oxygen atoms remained unchanged. Consequently, the carbon atoms in this reaction underwent oxidation, that is, they lost electrons ( oxidized), and the iron atoms – reduction, that is, they added electrons ( recovered) (see § 7.16). To characterize OVR, the concepts are used oxidizer And reducing agent.
Thus, in our reaction the oxidizing atoms are iron atoms, and the reducing atoms are carbon atoms.
In our reaction, the oxidizing agent is iron(III) oxide, and the reducing agent is carbon(II) monoxide.
In cases where oxidizing atoms and reducing atoms are part of the same substance (example: reaction 6 from the previous paragraph), the concepts of “oxidizing substance” and “reducing substance” are not used.
Thus, typical oxidizing agents are substances that contain atoms that tend to gain electrons (in whole or in part), lowering their oxidation state. Of the simple substances, these are primarily halogens and oxygen, and to a lesser extent sulfur and nitrogen. From complex substances - substances that contain atoms in higher oxidation states that are not inclined to form simple ions in these oxidation states: HNO 3 (N +V), KMnO 4 (Mn +VII), CrO 3 (Cr +VI), KClO 3 (Cl +V), KClO 4 (Cl +VII), etc.
Typical reducing agents are substances that contain atoms that tend to completely or partially donate electrons, increasing their oxidation state. Simple substances include hydrogen, alkali and alkaline earth metals, and aluminum. Of the complex substances - H 2 S and sulfides (S –II), SO 2 and sulfites (S +IV), iodides (I –I), CO (C +II), NH 3 (N –III), etc.
In general, almost all complex and many simple substances can exhibit both oxidizing and reducing properties. For example:
SO 2 + Cl 2 = S + Cl 2 O 2 (SO 2 is a strong reducing agent);
SO 2 + C = S + CO 2 (t) (SO 2 is a weak oxidizing agent);
C + O 2 = CO 2 (t) (C is a reducing agent);
C + 2Ca = Ca 2 C (t) (C is an oxidizing agent).
Let's return to the reaction we discussed at the beginning of this section.
| Fe2O3 | + | = | 2Fe | + |
Please note that as a result of the reaction, oxidizing atoms (Fe + III) turned into reducing atoms (Fe 0), and reducing atoms (C + II) turned into oxidizing atoms (C + IV). But CO 2 is a very weak oxidizing agent under any conditions, and iron, although it is a reducing agent, is under these conditions much weaker than CO. Therefore, the reaction products do not react with each other, and the reverse reaction does not occur. The given example is an illustration of the general principle that determines the direction of the flow of the OVR:
Redox reactions proceed in the direction of the formation of a weaker oxidizing agent and a weaker reducing agent.
The redox properties of substances can only be compared under identical conditions. In some cases, this comparison can be made quantitatively.
While doing your homework for the first paragraph of this chapter, you became convinced that it is quite difficult to select coefficients in some reaction equations (especially ORR). To simplify this task in the case of redox reactions, the following two methods are used:
A) electronic balance method And
b) electron-ion balance method.
You will learn the electron balance method now, and the electron-ion balance method is usually studied in higher education institutions.
Both of these methods are based on the fact that electrons in chemical reactions neither disappear nor appear anywhere, that is, the number of electrons accepted by atoms is equal to the number of electrons given up by other atoms.
The number of given and accepted electrons in the electron balance method is determined by the change in the oxidation state of atoms. When using this method, it is necessary to know the composition of both the starting substances and the reaction products.
Let's look at the application of the electronic balance method using examples.
Example 1. Let's create an equation for the reaction of iron with chlorine. It is known that the product of this reaction is iron(III) chloride. Let's write down the reaction scheme:
Fe + Cl 2 FeCl 3 .
Let us determine the oxidation states of the atoms of all elements that make up the substances participating in the reaction:
Iron atoms donate electrons, and chlorine molecules accept them. Let us express these processes electronic equations:
Fe – 3 e– = Fe +III,
Cl2+2 e –= 2Cl –I.
In order for the number of electrons given to be equal to the number of electrons received, the first electronic equation must be multiplied by two, and the second by three:
| Fe – 3 e– = Fe +III, Cl2+2 e– = 2Cl –I |
2Fe – 6 e– = 2Fe +III, 3Cl 2 + 6 e– = 6Cl –I. |
By introducing coefficients 2 and 3 into the reaction scheme, we obtain the reaction equation:
2Fe + 3Cl 2 = 2FeCl 3.
Example 2. Let's create an equation for the combustion reaction of white phosphorus in excess chlorine. It is known that phosphorus(V) chloride is formed under these conditions:
| +V –I | ||||
| P 4 | + | Cl2 | PCl 5. |
White phosphorus molecules give up electrons (oxidize), and chlorine molecules accept them (reduce):
| P 4 – 20 e– = 4P +V Cl2+2 e– = 2Cl –I |
1 10 |
2 20 |
P 4 – 20 e– = 4P +V Cl2+2 e– = 2Cl –I |
P 4 – 20 e– = 4P +V 10Cl 2 + 20 e– = 20Cl –I |
The initially obtained factors (2 and 20) had a common divisor, by which (like future coefficients in the reaction equation) they were divided. Reaction equation:
P4 + 10Cl2 = 4PCl5.
Example 3. Let's create an equation for the reaction that occurs when iron(II) sulfide is roasted in oxygen.
Reaction scheme:
| +III –II | +IV –II | |||||
| + | O2 | + |
In this case, both iron(II) and sulfur(–II) atoms are oxidized. The composition of iron(II) sulfide contains atoms of these elements in a 1:1 ratio (see the indices in the simplest formula).
Electronic balance:
| 4 | Fe+II – e– = Fe +III S–II–6 e– = S +IV |
In total they give 7 e – |
| 7 | O 2 + 4e – = 2O –II |
Reaction equation: 4FeS + 7O 2 = 2Fe 2 O 3 + 4SO 2.
Example 4. Let's create an equation for the reaction that occurs when iron(II) disulfide (pyrite) is roasted in oxygen.
Reaction scheme:
| +III –II | +IV –II | |||||
| + | O2 | + |
As in the previous example, both iron(II) atoms and sulfur atoms are also oxidized here, but with an oxidation state of I. The atoms of these elements are included in the composition of pyrite in a ratio of 1:2 (see the indices in the simplest formula). It is in this regard that the iron and sulfur atoms react, which is taken into account when compiling the electronic balance:
| Fe+III – e– = Fe +III 2S–I – 10 e– = 2S +IV |
In total they give 11 e – | |
| O2+4 e– = 2O –II |
Reaction equation: 4FeS 2 + 11O 2 = 2Fe 2 O 3 + 8SO 2.
There are also more complex cases of ODD, some of which you will become familiar with while doing your homework.
OXIDIZING ATOM, REDUCING ATOM, OXIDIZING SUBSTANCE, REDUCING SUBSTANCE, ELECTRONIC BALANCE METHOD, ELECTRONIC EQUATIONS.
1. Compile an electronic balance for each OVR equation given in the text of § 1 of this chapter.
2. Make up equations for the ORRs that you discovered while completing the task for § 1 of this chapter. This time, use the electronic balance method to set the odds. 3.Using the electron balance method, create reaction equations corresponding to the following schemes: a) Na + I 2 NaI;
b) Na + O 2 Na 2 O 2;
c) Na 2 O 2 + Na Na 2 O;
d) Al + Br 2 AlBr 3;
e) Fe + O 2 Fe 3 O 4 ( t);
e) Fe 3 O 4 + H 2 FeO + H 2 O ( t);
g) FeO + O 2 Fe 2 O 3 ( t);
i) Fe 2 O 3 + CO Fe + CO 2 ( t);
j) Cr + O 2 Cr 2 O 3 ( t);
l) CrO 3 + NH 3 Cr 2 O 3 + H 2 O + N 2 ( t);
m) Mn 2 O 7 + NH 3 MnO 2 + N 2 + H 2 O;
m) MnO 2 + H 2 Mn + H 2 O ( t);
n) MnS + O 2 MnO 2 + SO 2 ( t)
p) PbO 2 + CO Pb + CO 2 ( t);
c) Cu 2 O + Cu 2 S Cu + SO 2 ( t);
t) CuS + O 2 Cu 2 O +SO 2 ( t);
y) Pb 3 O 4 + H 2 Pb + H 2 O ( t).
9.3. Exothermic reactions. Enthalpy
Why do chemical reactions occur?
To answer this question, let us remember why individual atoms combine into molecules, why an ionic crystal is formed from isolated ions, and why the principle of least energy applies when the electron shell of an atom is formed. The answer to all these questions is the same: because it is energetically beneficial. This means that during such processes energy is released. It would seem that chemical reactions should occur for the same reason. Indeed, many reactions can be carried out, during which energy is released. Energy is released, usually in the form of heat.
If during an exothermic reaction the heat does not have time to be removed, then the reaction system heats up.
For example, in the methane combustion reaction
CH 4 (g) + 2O 2 (g) = CO 2 (g) + 2H 2 O (g)
so much heat is released that methane is used as fuel.
The fact that this reaction releases heat can be reflected in the reaction equation:
CH 4 (g) + 2O 2 (g) = CO 2 (g) + 2H 2 O (g) + Q.
This is the so called thermochemical equation. Here the symbol "+ Q" means that when methane is burned, heat is released. This heat is called thermal effect of reaction.
Where does the released heat come from?
You know that during chemical reactions chemical bonds are broken and formed. In this case, the bonds between carbon and hydrogen atoms in CH 4 molecules, as well as between oxygen atoms in O 2 molecules, are broken. In this case, new bonds are formed: between carbon and oxygen atoms in CO 2 molecules and between oxygen and hydrogen atoms in H 2 O molecules. To break bonds, you need to expend energy (see “bond energy”, “atomization energy”), and when forming bonds, energy is released. Obviously, if the “new” bonds are stronger than the “old” ones, then more energy will be released than absorbed. The difference between the released and absorbed energy is the thermal effect of the reaction.
Thermal effect (amount of heat) is measured in kilojoules, for example:
2H 2 (g) + O 2 (g) = 2H 2 O (g) + 484 kJ.
This notation means that 484 kilojoules of heat will be released if two moles of hydrogen react with one mole of oxygen to produce two moles of gaseous water (water vapor).
Thus, in thermochemical equations, the coefficients are numerically equal to the amounts of substance of the reactants and reaction products.
What determines the thermal effect of each specific reaction?
The thermal effect of the reaction depends
a) on the aggregative states of the starting substances and reaction products,
b) on temperature and
c) on whether the chemical transformation occurs at constant volume or at constant pressure.
The dependence of the thermal effect of a reaction on the state of aggregation of substances is due to the fact that the processes of transition from one state of aggregation to another (like some other physical processes) are accompanied by the release or absorption of heat. This can also be expressed by a thermochemical equation. Example – thermochemical equation for condensation of water vapor:
H 2 O (g) = H 2 O (l) + Q.
In thermochemical equations, and, if necessary, in ordinary chemical equations, the aggregative states of substances are indicated using letter indices:
(d) – gas,
(g) – liquid,
(t) or (cr) – solid or crystalline substance.
The dependence of the thermal effect on temperature is associated with differences in heat capacities
starting materials and reaction products.
Since the volume of the system always increases as a result of an exothermic reaction at constant pressure, part of the energy is spent on doing work to increase the volume, and the heat released will be less than if the same reaction occurs at a constant volume.
Thermal effects of reactions are usually calculated for reactions occurring at constant volume at 25 °C and are indicated by the symbol Q o.
If energy is released only in the form of heat, and a chemical reaction proceeds at a constant volume, then the thermal effect of the reaction ( Q V) is equal to the change internal energy
(D U) substances participating in the reaction, but with the opposite sign:
Q V = – U.
The internal energy of a body is understood as the total energy of intermolecular interactions, chemical bonds, the ionization energy of all electrons, the bond energy of nucleons in nuclei, and all other known and unknown types of energy “stored” by this body. The “–” sign is due to the fact that when heat is released, the internal energy decreases. That is
U= – Q V .
If the reaction occurs at constant pressure, then the volume of the system can change. Doing work to increase the volume also takes part of the internal energy. In this case
U = –(QP+A) = –(QP+PV),
Where Qp– the thermal effect of a reaction occurring at constant pressure. From here
Q P = – U–PV .
A value equal to U+PV got the name enthalpy change and denoted by D H.
H=U+PV.
Hence
Q P = – H.
Thus, as heat is released, the enthalpy of the system decreases. Hence the old name for this quantity: “heat content”.
Unlike the thermal effect, a change in enthalpy characterizes a reaction regardless of whether it occurs at constant volume or constant pressure. Thermochemical equations written using enthalpy change are called thermochemical equations in thermodynamic form. In this case, the value of the enthalpy change under standard conditions (25 °C, 101.3 kPa) is given, denoted H o. For example:
2H 2 (g) + O 2 (g) = 2H 2 O (g) H o= – 484 kJ;
CaO (cr) + H 2 O (l) = Ca(OH) 2 (cr) H o= – 65 kJ.
Dependence of the amount of heat released in the reaction ( Q) from the thermal effect of the reaction ( Q o) and the amount of substance ( n B) one of the participants in the reaction (substance B - the starting substance or reaction product) is expressed by the equation:
Here B is the amount of substance B, specified by the coefficient in front of the formula of substance B in the thermochemical equation.
Task
Determine the amount of hydrogen substance burned in oxygen if 1694 kJ of heat was released.
Solution
2H 2 (g) + O 2 (g) = 2H 2 O (g) + 484 kJ. |
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Q = 1694 kJ, 6. The thermal effect of the reaction between crystalline aluminum and gaseous chlorine is 1408 kJ. Write the thermochemical equation for this reaction and determine the mass of aluminum required to produce 2816 kJ of heat using this reaction. 9.4. Endothermic reactions. Entropy In addition to exothermic reactions, reactions are possible in which heat is absorbed, and if it is not supplied, the reaction system is cooled. Such reactions are called endothermic. The thermal effect of such reactions is negative. For example: Thus, the energy released during the formation of bonds in the products of these and similar reactions is less than the energy required to break bonds in the starting substances.
Let's take two flasks and fill one of them with nitrogen (colorless gas) and the other with nitrogen dioxide (brown gas) so that both the pressure and temperature in the flasks are the same. It is known that these substances do not react chemically with each other. Let's tightly connect the flasks with their necks and install them vertically, so that the flask with heavier nitrogen dioxide is at the bottom (Fig. 9.1). After some time, we will see that brown nitrogen dioxide gradually spreads into the upper flask, and colorless nitrogen penetrates into the lower one. As a result, the gases mix, and the color of the contents of the flasks becomes the same. Thus,
Equations of connection between entropy ( S) and other quantities are studied in physics and physical chemistry courses. Entropy unit [ S] = 1 J/K. G= H–T S Condition for spontaneous reaction: G< 0. At low temperatures, the factor determining the possibility of a reaction occurring is largely the energy factor, and at high temperatures it is the entropy factor. From the above equation, in particular, it is clear why decomposition reactions that do not occur at room temperature (entropy increases) begin to occur at elevated temperatures. ENDOTHERMIC REACTION, ENTROPY, ENERGY FACTOR, ENTROPY FACTOR, GIBBS ENERGY. 2CuO (cr) + C (graphite) = 2Cu (cr) + CO 2 (g) is –46 kJ. Write down the thermochemical equation and calculate how much energy is needed to produce 1 kg of copper from this reaction. CaCO 3 (cr) = CaO (cr) + CO 2 (g) – 179 kJ 24.6 liters of carbon dioxide were formed. Determine how much heat was wasted uselessly. How many grams of calcium oxide were formed? |
DEFINITION
Chemical reaction are called transformations of substances in which a change in their composition and (or) structure occurs.
Most often, chemical reactions are understood as the process of converting starting substances (reagents) into final substances (products).
Chemical reactions are written using chemical equations containing the formulas of the starting substances and reaction products. According to the law of conservation of mass, the number of atoms of each element on the left and right sides of a chemical equation is the same. Typically, the formulas of the starting substances are written on the left side of the equation, and the formulas of the products on the right. The equality of the number of atoms of each element on the left and right sides of the equation is achieved by placing integer stoichiometric coefficients in front of the formulas of substances.
Chemical equations may contain additional information about the characteristics of the reaction: temperature, pressure, radiation, etc., which is indicated by the corresponding symbol above (or “below”) the equal sign.
All chemical reactions can be grouped into several classes, which have certain characteristics.
Classification of chemical reactions according to the number and composition of starting and resulting substances
According to this classification, chemical reactions are divided into reactions of connection, decomposition, substitution, and exchange.
As a result compound reactions from two or more (complex or simple) substances one new substance is formed. IN general view The equation for such a chemical reaction will look like this:
For example:
CaCO 3 + CO 2 + H 2 O = Ca(HCO 3) 2
SO 3 + H 2 O = H 2 SO 4
2Mg + O 2 = 2MgO.
2FeCl 2 + Cl 2 = 2FeCl 3
The reactions of the compound are in most cases exothermic, i.e. proceed with the release of heat. If simple substances are involved in the reaction, then such reactions are most often redox reactions (ORR), i.e. occur with changes in the oxidation states of elements. It is clear to say whether there will be a connection reaction between complex substances cannot be treated as OVR.
Reactions that result in the formation of several other new substances (complex or simple) from one complex substance are classified as decomposition reactions. In general, the equation for the chemical reaction of decomposition will look like this:
For example:
CaCO 3 CaO + CO 2 (1)
2H 2 O = 2H 2 + O 2 (2)
CuSO 4 × 5H 2 O = CuSO 4 + 5H 2 O (3)
Cu(OH) 2 = CuO + H 2 O (4)
H 2 SiO 3 = SiO 2 + H 2 O (5)
2SO 3 =2SO 2 + O 2 (6)
(NH 4) 2 Cr 2 O 7 = Cr 2 O 3 + N 2 +4H 2 O (7)
Most decomposition reactions occur when heated (1,4,5). Possible decomposition due to exposure electric current(2). Decomposition of crystalline hydrates, acids, bases and salts oxygen-containing acids(1, 3, 4, 5, 7) occurs without changing the oxidation states of the elements, i.e. these reactions are not related to ODD. ORR decomposition reactions include the decomposition of oxides, acids and salts formed by elements in higher oxidation states (6).
Decomposition reactions also occur in organic chemistry, but under other names - cracking (8), dehydrogenation (9):
C 18 H 38 = C 9 H 18 + C 9 H 20 (8)
C 4 H 10 = C 4 H 6 + 2H 2 (9)
At substitution reactions a simple substance interacts with a complex substance, forming a new simple and a new complex substance. In general, the equation for a chemical substitution reaction will look like this:
For example:
2Al + Fe 2 O 3 = 2Fe + Al 2 O 3 (1)
Zn + 2HCl = ZnСl 2 + H 2 (2)
2KBr + Cl 2 = 2KCl + Br 2 (3)
2КlO 3 + l 2 = 2KlO 3 + Сl 2 (4)
CaCO 3 + SiO 2 = CaSiO 3 + CO 2 (5)
Ca 3 (PO 4) 2 + 3SiO 2 = 3СаSiO 3 + P 2 O 5 (6)
CH 4 + Cl 2 = CH 3 Cl + HCl (7)
Most substitution reactions are redox (1 – 4, 7). Examples of decomposition reactions in which no change in oxidation states occurs are few (5, 6).
Exchange reactions are reactions that occur between complex substances in which they exchange their constituent parts. Typically this term is used for reactions involving ions found in aqueous solution. In general, the equation for a chemical exchange reaction will look like this:
AB + CD = AD + CB
For example:
CuO + 2HCl = CuCl 2 + H 2 O (1)
NaOH + HCl = NaCl + H 2 O (2)
NaHCO 3 + HCl = NaCl + H 2 O + CO 2 (3)
AgNO 3 + KBr = AgBr ↓ + KNO 3 (4)
CrCl 3 + ZNaON = Cr(OH) 3 ↓+ ZNaCl (5)
Exchange reactions are not redox. A special case of these exchange reactions is the neutralization reaction (the reaction of acids with alkalis) (2). Exchange reactions proceed in the direction where at least one of the substances is removed from the reaction sphere in the form gaseous substance(3), sediment (4, 5) or a poorly dissociating compound, most often water (1, 2).
Classification of chemical reactions according to changes in oxidation states
Depending on the change in the oxidation states of the elements that make up the reagents and reaction products, all chemical reactions are divided into redox reactions (1, 2) and those occurring without changing the oxidation state (3, 4).
2Mg + CO 2 = 2MgO + C (1)
Mg 0 – 2e = Mg 2+ (reducing agent)
C 4+ + 4e = C 0 (oxidizing agent)
FeS 2 + 8HNO 3 (conc) = Fe(NO 3) 3 + 5NO + 2H 2 SO 4 + 2H 2 O (2)
Fe 2+ -e = Fe 3+ (reducing agent)
N 5+ +3e = N 2+ (oxidizing agent)
AgNO 3 +HCl = AgCl ↓ + HNO 3 (3)
Ca(OH) 2 + H 2 SO 4 = CaSO 4 ↓ + H 2 O (4)
Classification of chemical reactions by thermal effect
Depending on whether heat (energy) is released or absorbed during the reaction, all chemical reactions are conventionally divided into exothermic (1, 2) and endothermic (3), respectively. The amount of heat (energy) released or absorbed during a reaction is called the thermal effect of the reaction. If the equation indicates the amount of heat released or absorbed, then such equations are called thermochemical.
N 2 + 3H 2 = 2NH 3 +46.2 kJ (1)
2Mg + O 2 = 2MgO + 602.5 kJ (2)
N 2 + O 2 = 2NO – 90.4 kJ (3)
Classification of chemical reactions according to the direction of the reaction
Based on the direction of the reaction, reversible reactions are distinguished ( chemical processes, the products of which are capable of reacting with each other under the same conditions in which they were obtained, to form the starting substances) and irreversible (chemical processes whose products are not able to react with each other to form the starting substances).
For reversible reactions The equation in general form is usually written as follows:
A + B ↔ AB
For example:
CH 3 COOH + C 2 H 5 OH ↔ H 3 COOC 2 H 5 + H 2 O
Examples of irreversible reactions include the following reactions:
2КlО 3 → 2Кl + ЗО 2
C 6 H 12 O 6 + 6O 2 → 6CO 2 + 6H 2 O
Evidence of the irreversibility of a reaction can be the release of a gaseous substance, a precipitate, or a poorly dissociating compound, most often water, as reaction products.
Classification of chemical reactions according to the presence of a catalyst
From this point of view, catalytic and non-catalytic reactions are distinguished.
A catalyst is a substance that speeds up the progress of a chemical reaction. Reactions that occur with the participation of catalysts are called catalytic. Some reactions cannot take place at all without the presence of a catalyst:
2H 2 O 2 = 2H 2 O + O 2 (MnO 2 catalyst)
Often one of the reaction products serves as a catalyst that accelerates this reaction (autocatalytic reactions):
MeO+ 2HF = MeF 2 + H 2 O, where Me is a metal.
Examples of problem solving
EXAMPLE 1
Let's talk about how to create a chemical equation, because they are the main elements of this discipline. Thanks to a deep understanding of all the patterns of interactions and substances, you can control them, apply them in various fields activities.
Theoretical features
Drawing up chemical equations is an important and responsible stage, considered in the eighth grade. secondary schools. What should precede this stage? Before the teacher tells his students how to create a chemical equation, it is important to introduce schoolchildren to the term “valence” and teach them to determine this value for metals and non-metals using the periodic table of elements.
Compilation of binary formulas by valence
In order to understand how to create a chemical equation by valency, you first need to learn how to create formulas for compounds consisting of two elements using valence. We propose an algorithm that will help cope with the task. For example, you need to create a formula for sodium oxide.
First, it is important to take into account that the chemical element that is mentioned last in the name should be in first place in the formula. In our case, sodium will be written first in the formula, oxygen second. Let us recall that oxides are binary compounds in which the last (second) element must be oxygen with an oxidation state of -2 (valency 2). Next, using the periodic table, it is necessary to determine the valence of each of the two elements. To do this we use certain rules.
Since sodium is a metal that is located in main subgroup 1 group, its valence is a constant value, it is equal to I.
Oxygen is a non-metal, since it is the last one in the oxide; to determine its valence, we subtract 6 from eight (the number of groups) (the group in which oxygen is located), we obtain that the valency of oxygen is II.
Between certain valences we find the least common multiple, then divide it by the valency of each of the elements to obtain their indices. We write down the finished formula Na 2 O.
Instructions for composing an equation
Now let's talk in more detail about how to write a chemical equation. First, let's look at the theoretical aspects, then move on to specific examples. So, composing chemical equations presupposes a certain procedure.
- 1st stage. After reading the proposed task, you need to determine which chemical substances must be present on the left side of the equation. A “+” sign is placed between the original components.
- 2nd stage. After the equal sign, you need to create a formula for the reaction product. When performing such actions, you will need the algorithm for composing formulas for binary compounds, which we discussed above.
- 3rd stage. We check the number of atoms of each element before and after chemical interaction, if necessary, we put additional coefficients in front of the formulas.
Example of a combustion reaction
Let's try to figure out how to create a chemical equation for the combustion of magnesium using an algorithm. On the left side of the equation we write the sum of magnesium and oxygen. Do not forget that oxygen is a diatomic molecule, so it must be given an index of 2. After the equal sign, we compose the formula for the product obtained after the reaction. It will be in which magnesium is written first, and oxygen is written second in the formula. Next, using the table of chemical elements, we determine the valencies. Magnesium, which is in group 2 (the main subgroup), has a constant valency II; for oxygen, by subtracting 8 - 6 we also get valency II.
The process record will look like: Mg+O 2 =MgO.
In order for the equation to comply with the law of conservation of mass of substances, it is necessary to arrange the coefficients. First, we check the amount of oxygen before the reaction, after the process is completed. Since there were 2 oxygen atoms, but only one was formed, a coefficient of 2 must be added on the right side before the magnesium oxide formula. Next, we count the number of magnesium atoms before and after the process. As a result of the interaction, 2 magnesium was obtained, therefore, on the left side in front of the simple substance magnesium, a coefficient of 2 is also required.
The final type of reaction: 2Mg+O 2 =2MgO.
Example of a substitution reaction
Any chemistry abstract contains a description different types interactions.
Unlike a compound, in a substitution there will be two substances on both the left and right sides of the equation. Let's say we need to write the reaction of interaction between zinc and We use the standard writing algorithm. First, on the left side we write zinc and hydrochloric acid through the sum, and on the right side we write the formulas for the resulting reaction products. Since zinc is located before hydrogen in the electrochemical voltage series of metals, in this process it displaces molecular hydrogen from the acid and forms zinc chloride. As a result, we get the following entry: Zn+HCL=ZnCl 2 +H 2.
Now we move on to equalizing the number of atoms of each element. Since there was one atom on the left side of chlorine, and after the interaction there were two, before the formula of hydrochloric acid it is necessary to set a coefficient of 2.
As a result, we obtain a ready-made reaction equation corresponding to the law of conservation of mass of substances: Zn+2HCL=ZnCl 2 +H 2 .
Conclusion
A typical chemistry note necessarily contains several chemical transformations. Not a single section of this science is limited to simple verbal description transformations, dissolution processes, evaporation, everything is necessarily confirmed by equations. The specificity of chemistry lies in the fact that all processes that occur between different inorganic or organic substances, can be described using coefficients and indices.
How else does chemistry differ from other sciences? Chemical equations help not only to describe the transformations that occur, but also to carry out quantitative calculations on them, thanks to which it is possible to carry out laboratory and industrial production different substances.